How Accurate Is the Calculation of Potential Energy Change at High Altitudes?

AI Thread Summary
The discussion focuses on calculating the potential energy change for an 80kg mountaineer ascending Mount Everest, which is 9km high. The initial calculation using ΔEp = mgΔh yields approximately 7.1 million joules, but participants note that this is an estimate since the gravitational acceleration (g) changes with altitude. A more precise calculation involves integrating Newton's law of gravitation to account for the varying value of g at different heights. The integration leads to the formula ΔU = GM_E m (1/R - 1/(R+h)), allowing for a more accurate determination of potential energy change. This approach emphasizes the importance of considering gravitational variation when calculating potential energy at high altitudes.
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Homework Statement



Calculate the gain of potential energy of a mountaineer of mass 80kg who travels to the top of the mountain Everest 9km above sea level.


Homework Equations



ΔEp=mgΔh

The Attempt at a Solution



ΔEp=mgΔh=80x9.81x9x10^3=7.1x10^6J.

This is the correct answer but surely it's only an estimate since the value of g 9km above the Earth's surface is 9.83830. Δg=9.8380-9.81= 2.8x10^(-3)?

Is it possible to calculate the exact change is Ep as the mass moves through Δh, rather than using a fixed value of 9.81?
 
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studentxlol said:

Homework Statement



Calculate the gain of potential energy of a mountaineer of mass 80kg who travels to the top of the mountain Everest 9km above sea level.


Homework Equations



ΔEp=mgΔh

The Attempt at a Solution



ΔEp=mgΔh=80x9.81x9x10^3=7.1x10^6J.

This is the correct answer but surely it's only an estimate since the value of g 9km above the Earth's surface is 9.83830. Δg=9.8380-9.81= 2.8x10^(-3)?

Is it possible to calculate the exact change is Ep as the mass moves through Δh, rather than using a fixed value of 9.81?
Yes. One would need to go back to Newton's law of gravitation and compute

\Delta U = \int_{R}^{R+h} \frac{GM_E m}{r^2}\;\text{d}r,

where R is the radius of the Earth, ME is the mass of the Earth, m is the mass of the man and h is the height of Everest.
 
Hootenanny said:
Yes. One would need to go back to Newton's law of gravitation and compute

\Delta U = \int_{R}^{R+h} \frac{GM_E m}{r^2}\;\text{d}r,

where R is the radius of the Earth, ME is the mass of the Earth, m is the mass of the man and h is the height of Everest.

Ah ok.

How would you go about integrating GMm/r^2? My maths isn't great but this is what I get:

Since G is a constant, factor it out.

\Delta U = G\int_{R}^{R+h} \frac{M_E m}{r^2}\;\text{d}r,

\Delta U = G [-\frac{M_E m}{r}]_{R}^{R+h}

Sorry, my maths isn't at this level yet!
 
studentxlol said:
Ah ok.

How would you go about integrating GMm/r^2? My maths isn't great but this is what I get:

Since G is a constant, factor it out.

\Delta U = G\int_{R}^{R+h} \frac{M_E m}{r^2}\;\text{d}r,

\Delta U = G [-\frac{M_E m}{r}]_{R}^{R+h}

Sorry, my maths isn't at this level yet!
So we are left with
\Delta U = GM_E m \left(\frac{1}{R} - \frac{1}{R+h}\right).
Simply plug the numbers in and you should get your result.
 
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